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CURVE TABLES. <br /> Published by KEUFFEL Bic EMR CA. <br /> HOW TO USE CURVE TABLES. <br /> Table I. contains'Tangents and Externals to a V curve. Tan.and <br /> t.to any other radius maybe found nearly enough,by dividing the Tan. <br /> ! F Ext.opposite the given Central Angle by thegiven degree of curve. <br /> To find Deg. of Curve, having the Central Angle and Tangent: <br /> y ivide Tan.opposite the given Central Angle by the given Tangent. <br /> I E To find Deg. of Curve, having the Central Angle and External: <br /> Divide Eat. opposite the gI'ven Central Angle by the given External. <br /> To find Nat.Tam.and Nat.Ex.See.for any angle by Table I.:Tan. <br /> Ext.of twice the given angle divided by the radius of a V curve will <br /> the Nat.Tan.or Nat.Ex.See. <br /> EXAMPLE. <br /> Wanted a Curve with an Ext.of about 12 ft. Angle <br /> of Intersection or L P.-23° 20' to the R. at Station <br /> S^ 542+72. <br /> Ext.in Tab. I opposite 23°20'=120.87 <br /> 0 v i v 120.87+12=10.07. Say a 10°Curve. <br /> Tan.in Tab: I opp.23°20'=1183.1 <br /> 1183.1-1~10=118.31. <br /> /4L ' Correction for A.23°20'for a 10°Cur.—0.16 <br /> -- <br /> 118.31+0.16=118.47=corrected Tangent. <br /> (If corrected Ext.is required find in same way) <br /> Ang.23°20'=23.33 +10=2.3333=L.C. <br /> 20 191'=def.for sta. 542 I,P.=sta. 542+72 <br /> 2 off. 4°49#'= " +50 Tan.= 1 .18.47 <br /> 7"19,}'= " " 543 <br /> 9 �l 9'49j'= " " +50 B.C.=sta. 541+53.53 <br /> 110 401= " " " 543- L.C.= 2 .33.33 <br /> ! — 86.86 E.C.=Sta. 543+86.86 <br /> 100-53.53=46.47X31(def.for 1 ft.of 10°Cur.)=139.41'= <br /> 20 191'=def.for sta.542. <br /> / - Def,for 50 ft.=2°30'fora 10°Curve. <br /> Def.for 36.86 ft.=1°50}'for a 10*Curve. <br /> e� <br /> tx <br /> LP.A+�.Y3•YO <br /> ry <br /> 5101 Curve <br /> 47 <br /> \ j <br /> i <br />